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Consider the family of ellipses defined by $\frac{u^2}{(c+\frac{1}{c})^2}+\frac{v^2}{(c-\frac{1}{c})^2}=1$ with $0<c<1$. They are the image of the circles $r=c$ under the mapping $w=z+\frac{1}{z}$. I can show that no two of the ellipses have a point in common. But how would one show that they cover the whole plane, except for the bit on the $u$-axis between $-2$ and $2$ ? I thought of working with the double valued "inverse" $z=\frac{1}{2}(w + (w^2-4)^{\frac{1}{2}})$ but this would involve showing that the absolute value of one those values is less than $1$, which looks a rather lengthy calculation.

dario
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Let's start with the classical studies of functions $f:x\mapsto x+\frac1x$ and $g:x\mapsto x-\frac1x$ on $(0,1)$.

$\forall x\in (0,1), f(x)>2$ and $f$ decreases continuously from $+\infty$ to $2^{-}$; at the same time, $g$ grows continuously from $-\infty$ to $0^{-}$.

$\forall c\in (0,1), f(c) $ and $-g(c)$ represent respectively the semi-major and semi-minor axes of the ellipses.

So the family of ellipses covers the plane, except for the bit on the u-axis between −2 and 2, which is illustrated by the graph proposed by @Donald Splutterwit.

Stéphane Jaouen
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  • Yes, that was what I did to prove that no two ellipses have a point in common, but does it really prove that the plane is covered? I don't see how. – dario Oct 28 '22 at 17:02
  • It's true. I don't feel like giving a formal proof of it, but the argument: $(0,1)\to(\mathbb R^{+})^2,x\mapsto (f(x),g(x))$ is continuous, seemed quite significant to me. Otherwise, simply revert to the fact that the set of circles $r=c$ covers $\mathbb C^{*}$ and to $z\mapsto z+\frac{1}{z}$; We are back to the starting point of the post. :) – Stéphane Jaouen Oct 28 '22 at 17:18
  • I'm thinking about it: maybe : let $z_0 \in \mathbb C$. Let's proof that $\exists z \in \mathbb C : z_0=z+\frac{1}{z}$; this is equivalent to $z_0=\frac{z^2+1}{z} \iff z^2-z_0z+1=0$; this equation has a solution; this is good, isn't it ? – Stéphane Jaouen Oct 28 '22 at 17:31